Clustering and classification

Introduction to the data and data wrangling

In this chapter, we study Boston housing data from R package MASS, which contains records on housing values in suburbs of Boston. The data includes observations from 506 suburbs and has 14 variables. Variable definitions can be found in here.

str(boston)

## Classes 'tbl_df', 'tbl' and 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

Below is the summary of the variables in the data. There are 12 numeric and 2 integer variables in the data set and these all have different scale ranges.

summary(boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

The instructions of this exercise suggested to present a graphical overview of the data and to describe the outputs and comment the distributions of the variables and their relationships. The graphical overview of the data gets very difficult to read because there are 14 variables in the data; plotting the variables pairwise leads to very small and hard to read output if functions introduced in the course are used (i.e. GGAlly::ggpairs or pairs).

Therefore, a new function corrplot::corrplot is introduced. The correlation plot below shows that several of the 14 variables are positively or negatively correlated to varying degrees.

cor_matrix <- cor(boston)

corrplot(cor_matrix, method = "circle", type = "upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)

Because all the variables have different scales, let’s scale the data so that all the variables have mean value of 0 and standard deviation of 1.

boston_scaled <- as_tibble(scale(boston))

summary(boston_scaled)

##       crim                 zn               indus        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202  
##       chas              nox                rm               age         
##  Min.   :-0.2723   Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331  
##  1st Qu.:-0.2723   1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366  
##  Median :-0.2723   Median :-0.1441   Median :-0.1084   Median : 0.3171  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.:-0.2723   3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059  
##  Max.   : 3.6648   Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164  
##       dis               rad               tax             ptratio       
##  Min.   :-1.2658   Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047  
##  1st Qu.:-0.8049   1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876  
##  Median :-0.2790   Median :-0.5225   Median :-0.4642   Median : 0.2746  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6617   3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058  
##  Max.   : 3.9566   Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372  
##      black             lstat              medv        
##  Min.   :-3.9033   Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.: 0.2049   1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median : 0.3808   Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.4332   3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 0.4406   Max.   : 3.5453   Max.   : 2.9865

Some data wrangling before moving on to linear discriminant analysis:

# Creating a categorical crime variable

bins <- quantile(boston_scaled$crim)
crim_labels <- c("low", "med_low", "med_high", "high")

boston_scaled <- boston_scaled %>%
  mutate(crim = 
           cut(crim,
               breaks = bins,
               labels = crim_labels,
               include.lowest = TRUE))

# splitting the data to traning and test sets

n <- nrow(boston_scaled)
ind <- sample(n,  size = n * 0.8)

train <- boston_scaled[ind,]
test <- boston_scaled[-ind,]

Correct <- test$crim
test <- dplyr::select(test, -crim)

Linear discriminant analysis (LDA)

Fitting and printing the LDA model.

# fitting the model
lda.fit <- lda(crim ~ ., data = train)
lda.fit

## Call:
## lda(crim ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2425743 0.2574257 0.2475248 0.2524752 
## 
## Group means:
##                  zn      indus        chas        nox          rm
## low       1.0545866 -0.8850980 -0.19198008 -0.8858934  0.41378575
## med_low  -0.1174249 -0.2679064 -0.04518867 -0.5218771 -0.15989636
## med_high -0.3826200  0.2428040  0.23949396  0.4255776  0.04765841
## high     -0.4872402  1.0171096 -0.11793298  1.0568196 -0.37258050
##                 age        dis        rad        tax     ptratio
## low      -0.8658197  0.9457671 -0.7111611 -0.7323065 -0.44042345
## med_low  -0.2453951  0.2829744 -0.5434660 -0.4629581 -0.07050939
## med_high  0.4904674 -0.3769060 -0.4489825 -0.3157593 -0.29170877
## high      0.8120205 -0.8466221  1.6382099  1.5141140  0.78087177
##                black       lstat        medv
## low       0.37608626 -0.76709312  0.44045029
## med_low   0.31381417 -0.06289274 -0.03127224
## med_high  0.03911029  0.08496451  0.10690253
## high     -0.65243466  0.87626006 -0.74069164
## 
## Coefficients of linear discriminants:
##                 LD1          LD2         LD3
## zn       0.09550426  0.768950043 -0.97716375
## indus    0.06064181 -0.157486881  0.09180886
## chas    -0.09107653 -0.173319224  0.05020740
## nox      0.27327906 -0.668773997 -1.41579799
## rm      -0.12292904 -0.030816850 -0.24562690
## age      0.24233001 -0.315888720 -0.30287230
## dis     -0.05172074 -0.308928202 -0.04804057
## rad      3.48459781  0.990780887 -0.06502463
## tax     -0.02515273 -0.166546745  0.67885199
## ptratio  0.09208819  0.009619889 -0.31676846
## black   -0.09107938  0.061537554  0.13437331
## lstat    0.23849254 -0.362937810  0.44569787
## medv     0.20012145 -0.523372912 -0.06269099
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9474 0.0399 0.0127

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "grey", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(train$crim)

# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes, main ="LDA (bi)plot")
lda.arrows(lda.fit, myscale = 20)

Using test data set to predict the classes with the LDA model.

The crosstable below shows how well the LDA model predicts the crime rate in Boston suburbs. The LDA model is very accurate in predicting suburbs with high crime. Additionally, concerning suburbs with actual crime rate from low to medium high, the LDA model predicts crime rate rather accurately in the same region even if the prediction is not as accurate as with suburbs with high crime rate.

# fitting the model 
lda.predict <- predict(lda.fit, newdata = test)

# predicted values
Predicted <- lda.predict$class

# table of predicted vs. correct classes
descr::crosstab(Correct, Predicted, prop.r = T, plot = F)

##    Cell Contents 
## |-------------------------|
## |                   Count | 
## |             Row Percent | 
## |-------------------------|
## 
## =======================================================
##             Predicted
## Correct       low   med_low   med_high     high   Total
## -------------------------------------------------------
## low           13        15          1        0      29 
##             44.8%     51.7%       3.4%     0.0%   28.4%
## -------------------------------------------------------
## med_low        5        16          1        0      22 
##             22.7%     72.7%       4.5%     0.0%   21.6%
## -------------------------------------------------------
## med_high       0         7         17        2      26 
##              0.0%     26.9%      65.4%     7.7%   25.5%
## -------------------------------------------------------
## high           0         0          0       25      25 
##              0.0%      0.0%       0.0%   100.0%   24.5%
## -------------------------------------------------------
## Total         18        38         19       27     102 
## =======================================================

K-means clustering

Below is the summary of euclidian distances between observations.

# the data
boston <- as_tibble(MASS::Boston)
boston_scaled <- as_tibble(scale(boston))

# distance matrix 
dist_eu <- dist(boston_scaled, method = "euclidian")
summary(dist_eu)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

Here is the first set of k-means clustering done with four clusters. For easier inspection, I only plotted the pairs of firts six variables in the data set.

km <- kmeans(boston_scaled, centers = 4)

# plot the Boston dataset with clusters
pairs(boston_scaled[1:6], col = km$cluster)

I investigated the correct number of cluster by checking the total of within cluster sum of squares. After determining that the optimal number of clusters is two, I plotted the clusters again.

Interpratetion: the clusters seem to vary in across several variables. Looking at the crime variable, which has been the center if interest in this exercise, it looks like that the other cluster (black), includes suburbs with higher crime rates. The other cluster, however, includes only low crime rate suburbs.

# MASS, ggplot2 and Boston dataset are available
set.seed(123)

# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

# k-means clustering
km <-kmeans(boston_scaled, centers = 2)

# plot the Boston dataset with clusters
pairs(boston_scaled[1:6], col = km$cluster)

Super bonus

Preparing the data.

model_predictors <- dplyr::select(train, -crim)

# check the dimensions
dim(model_predictors)

## [1] 404  13

dim(lda.fit$scaling)

## [1] 13  3

The figure below plots the fitted LDA model in three dimensions.

# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type = 'scatter3d', mode = 'markers', colors = "Dark2")

Here is the same plot with the colors of the points representing crime rates of Boston suburbs.

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type = 'scatter3d', mode = 'markers', color = train$crim, colors = "Dark2")

Lastly, I did the same plot but this time the points are colored by using the two categories found in the k-means clustering exercise. Comparing the two last plots visually shows that k-means clustering identified those suburbs (green dots) that had high or medium high crime rate in the previous plot. This works as a one way to validate the k-means clustering categories.

km <-kmeans(boston_scaled[2:14], centers = 2)
clusters <- factor(km$cluster[ind])

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type = 'scatter3d', mode = 'markers', color = clusters, colors = "Dark2")